Algebra Review by Charles Denlinger and Elaine Jacobson (Auth.)

By Charles Denlinger and Elaine Jacobson (Auth.)

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56. EXAMPLE 2 Solve 3x - 2(2x - 5) = 2(x + 3) SOLUTION 3x - 4x + 10 = 2x + 6 - 8 - (removed parentheses) x + 10 = 2x - 2 (simplified) 10 = 3x - 2 (added x to both sides) 12 = 3x (added 2 to both sides) 4 = x EXAMPLE 3 SOLUTION C 1 2X Solve: (divided both sides by 3) + 5 , X - 4 ry r + — r =2 . To clear the equation of fractions, we multiply both sides by 6. - (6)<2) 3(2x + 5) + 2(x - 4) = 12 6x + 15 + 2x - 8 = 12 8x + 7 = 12 8x = x EXAMPLE 4 SOLUTION 5 8 Solve 3(2x - 3) = (x + 2)5 6x - 9 = 5x + 10 x - 9 =10 x = 19 (multiplied both sides by 3(x + 2) 57.

19. 20. 21. 22. 23. 24. 25. 26. 27. 28. 29. 50. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 40. i . [i - ( i . (i 51. 1 INTRODUCTION If 2 is substituted in the equation 3x- 2 = 4 the two sides become equal. We say that x = 2 is a solution of the equation. The following operations are helpful for finding the solu­ tions of an equation: I. II. The same quantity may be added to or subtracted from both sides of an equation. Both sides of an equation may be multiplied (or divided) by the same non-zero quantity.

3 Factor each of the following: 1. 22ab + 4b 3 2. ,32 02^0 2 4x y - 2x y + 8xy 3. 2x(x-l) + 3(x-l) 4. 1 + x+ 5. x - 8x+16 6. x + x - 12 7. 2 x - 9x- 5 8. 3x - 2x - 3 9. 3x -llx+6 10. x -llxy+24y 11. 2 3 x y-y 12. 2 2 16a - 9 b 13. 4a +4a- 3 14. x - 13x - 30 15. 3 2 2x + 5x^ - 3x 16. 2x - 8x 17. 6x - x - 5x 18. 8x - 2x - 3x 19. 2x + 5x - 3 20. x - 13x + 36 21. 3 3 xy - x y 22. 6x - 7x - 3 23. 9x + 3 0 x + 25 24. x -81 25. x -2 2 2 2 2 J 4 2 2 2 X 2 + 3 X 2 2 2 Z 2 4 4 4 3 2 2 4 2 2 2 37. 4 THE BINOMIAL THEOREM The array of numbers 1 1 1 1 1 1 3 10 6 15 1 1 3 6 4 5 1 2 1 20 1 4 10 5 15 6 1 1 is known as Pascal's triangle.

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